Question:
A star initially has $10^{40}$ deuterons. It produces energy via the
processes $_1H^2+_1H^2 \rightarrow _1H^3+p$ and $_1H^2+_1H^3 \rightarrow _2{He}^3+n$.
If the average power radiated by the star is $10^{16}$ W, the
deuteron supply of the star is exhausted in a time o f the order of
$M(H)^2=2.014 \, amu;$ $M(n)=1.008 \, amu;$
$M(p)=1.007 \, amu;$ $M(He)^2=4.001 \, amu;$
A star initially has $10^{40}$ deuterons. It produces energy via the
processes $_1H^2+_1H^2 \rightarrow _1H^3+p$ and $_1H^2+_1H^3 \rightarrow _2{He}^3+n$.
If the average power radiated by the star is $10^{16}$ W, the
deuteron supply of the star is exhausted in a time o f the order of
$M(H)^2=2.014 \, amu;$ $M(n)=1.008 \, amu;$
$M(p)=1.007 \, amu;$ $M(He)^2=4.001 \, amu;$
Updated On: Aug 15, 2022
- ${10}^6$s
- ${10}^8$s
- ${10}^{12}$s
- ${10}^{16}$s
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The Correct Option is C
Solution and Explanation
The given reactions are :
$_1H^2+_1H^2 \rightarrow _1H^3+p$
$_1H^2+_1H^3 \rightarrow _2{He}^4+n$.
$3_1H^2 \rightarrow _2{He}^4+n+p$.
$\triangle m=(3 \times 2.014- 4.001 - 1.007 - 1.008) amu$
= 0.026 amu
Energy released =$=0.026 \times 931\, MeV$
$=0.026 \times 931 \times 1.6 \times {10}^{-13}J$
$= 3.87 \times {10}^{-12}J$
This is the energy produced by the consumption of three
deuteron atoms.
$\therefore$ Total energy released by ${10}^{40}$ deuterons
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Following are the terms related to nucleus:
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